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Table 2 The CBSN numerical results for Example 4.1 with initial \(\pmb{z_{(1)}^{0}}\)

From: A generalized Newton method of high-order convergence for solving the large-scale linear complementarity problem

  

The performance of numerical results

The solution pairs \(\boldsymbol {(u^{*},w^{*})}\)

n = 3

It

4

u∗ = (0.3571,0.4286,0.3571)

CPU

0.0104

 

RES

0

w∗ = 1.0e − 011∗(0.8880,−0.5869,0.8880)

n = 5

It

4

u∗ = (0.3654,0.4615,0.4808,0.4615,0.3654)

CPU

0.0106

 

RES

6.1063e − 016

w∗ = 1.0e − 015∗(0,0,0,0,−0.1110)

n = 8

It

4

\(\begin{array}{lcl} u*&=&(0.3660,0.4641,0.4902,0.4967,0.4967,\\ &&{}0.4902,0.4641,0.3660) \end{array}\)

CPU

0.0113

 

RES

6.2042e − 016

\(\begin{array}{lcl} w*&=&1.0e\!-\!011*(0.4839,-0.1609,-0.0001,0,0,\\ &&{}-0.0001,-0.1608,0.4839) \end{array}\)

n = 10

It

3

\(\begin{array}{lcl} u*&=&(0.3660,0.4641,0.4904,0.4974, 0.4991,\\ &&{}0.4991,0.4974,0.4904,0.4641,0.3660) \end{array}\)

CPU

0.0115

 

RES

6.2232e − 016

\(\begin{array}{lcl} w*&=&1.0e\!-\!011*(0.4835,-0.1607,-0.0001,0,0,\\ &&{}0,0,-0.0001,-0.1607,0.4835) \end{array}\)